HDU 2066

/*ACMer:MDK
Accepted 2066 46MS 4284K 1465 B G++ MDK 
FOR:hdu 2066*/
#include<iostream>
#include<string.h>
#include<limits.h>
#include<stdio.h>
#define MAXN 1005
#define INTMAX 1000000 //这里取的大小让我纠结了很长时间我一开始去0x7fffffff，居然有负数，不知道为啥
using namespace std;
int map[MAXN][MAXN],minl[MAXN],pre[MAXN],live[100],goal[100];
int big = INT_MIN;
int dijkstra(int T,int S,int D)
{
    int n = ++big,i,j,k;
    int v[MAXN];
    for(i = 0;i<n;i++)
        v[i]=0,minl[i]=INTMAX,pre[i] = -1;
    for(minl[0]=0,j = 0;j<n;j++)
    {
        for(k = -1,i=0;i<n;i++)
            if(!v[i]&&(k==-1||minl[i]<minl[k]))
                k=i;
        for(v[k]=1,i=0;i<n;i++)
            if(!v[i]&&map[k][j]!=INTMAX&&minl[k]+map[k][i]<minl[i])
                minl[i]=minl[k]+map[pre[i]=k][i];//,cout<<minl[i]<<endl
    }
    int mina = INT_MAX;
    for(i = 0;i<D;i++)
        mina = min(mina,minl[goal[i]]);//,cout<<minl[goal[i]]<<endl
    return mina;
}
int main()
{
    freopen("d:\\1.txt","r",stdin);
    int T,S,D;
    while(cin>>T>>S>>D&&T)
    {
        int a,b;
        memset(map,INTMAX,sizeof(map));
        memset(minl,INTMAX,sizeof(minl));
        memset(pre,0,sizeof(pre));
        for(int i = 0,v;i<T;i++){
            cin>>a>>b>>v,big = max(big,max(a,b));
            if( v < map[a][b] ) map[a][b] = map[b][a] = v;
        }//,cout<<map[b][a]<<endl
        for(int i = 0;i<S;i++)
            cin>>live[i],map[0][live[i]]=map[live[i]][0]=0;
        for(int i = 0;i<D;i++)
            cin>>goal[i];
        cout<<dijkstra(T,S,D)<<endl;
    }
}

一开始用Floyd算法，果断超时，后来想想dijkstra还是可行的，近似的可以用多源，输入要加个判断，测试数据很是强烈。